Cantor Confusion
in [Math]
|
Prev: Pi berechnen: Ramanujan oder BBP
Next: Group Theory
From: mueckenh on 19 Nov 2006 07:34 Franziska Neugebauer schrieb: > Virgil wrote: > > > In article <1163856355.707119.306700(a)m7g2000cwm.googlegroups.com>, > > mueckenh(a)rz.fh-augsburg.de wrote: > > > >> Virgil schrieb: > [...] > >> And there are lines as long as the diagonal is. > > > > Name one. The elements of the diagonal are a subset of the line ends. Regards, WM
From: William Hughes on 19 Nov 2006 07:35 mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > William Hughes schrieb: > > > > > > > > I claim case iia: There is a potentially infinite sequence N = > > > > > 1,2,3,..., such that for any n there is n+1 but we cannot recognize or > > > > > treat all of its elements. In particular we can never complete this > > > > > set. We can never put it into a list > > > > > > > > OK. Knock yourself out. > > > > > > I read this several times from you. What does it mean? > > > > > > > Note however that this case > > > > is not consistent with assuming the axiom of infinity. > > > > The axiom of infinity says that the set N exists. > > > > > > But it does not say that it has an ordinal number and a cardinal > > > number. My case iia is in complete agreement with the axiom of > > > infinity. > > > > > > > > > The axiom of infinity says that the set N exists. Your iia says > > > > "we cannot recognize or treat all of its elements" > > This is no contradiction to the axiom. Compare the proof that the real > numbers can be well-ordered. We cannot construct or define or > recognize a well ordering. The fact that we cannot construct a real ordering does not stop us from proving such an ordering exists and using the existence of such an ordering. > > > > The axiom of infinity does not say anything about ordinal or > > cardinal numbers. However, given that the set N exists and > > the defnition of ordinal and cardinal numbers, it is easy to > > see that if N exists it must have both an ordinal and a cardinal > > number. > > > No. You assume the possibility of a bijection of the set with itself. > That is not proven from the mere existence of the set if we cannot > recognize or treat all of its elements. Piffle >And even if we could, the > axiom of infinity does not prove that infinite ordinal and cardinal > numbers are numbers, i.e., that they stand in trichotomy with each > other. Check the definitions of ordinal and cardinal below. Look for the term trichotomy. Fail to find the term trichotomy. Draw the obvious conclusion. > > > Ordinal. > > > > Every natural number is an ordinal. > > Every initial sequence of ordinals is an ordinal. > > (an initial sequence is a set of ordinals, such that > > if a is in the set every ordinal less than a is in the set). > > The set of all natural numbers is an initial sequence of > > ordinals. Therefore the set of natural numbers is an > > ordinal. > > > > Cardinal. > > > > Every natural number is a cardinal number too. > > > Cardinal "numbers" are equivalence classes of > > sets under the equivalence relation bijection. Since any > > set has a bijection to itself, > > Why? There are even models without a bijection to the natural numbers. Piffle. If the natural numbers exist, then the identity map is a bijection. - William Hughes
From: mueckenh on 19 Nov 2006 07:45 William Hughes schrieb: > > > > > > > > > > > > 1 > > > > > > 2 > > > > > > 3 > > > > > > ... > > > > > > n <--> 1,2,3,...n > > > > > > > > > > Please distinguish: > > > > iia: There is no number counting the elements of N. > > > > iib: There is a number omega counting the elements of N. > > > > No. You are trying to show that assuming case iia leads > to a contradiction. No. Case iia does not lead to a contradiction. > To do this you need to assume case iia. In > particular > you are assuming > > - the set, N, of all natural numbers exists > - the set N is infinite. > - N has no last element Yes. > > The diagonal contains all the d_nn. No line contains all the d_nn. > Therefore the diagonal is longer than every line. The diagonal consists of line indexes, i.e., of the line ends. Therefore it is a subset of the line indexes. > > The set of all lines contains every d_nn. No single line > contains every d_nn. The diagonal contains all d_nn. > The diagonal is longer than every line. The diagonal consists of line indexes, i.e., of the line ends. Therefore it is a subset of the line indexes. > > > > So we have two results: > > 1) The diagonal must be longer than every line. > > 2) The diagonal cannot be longer than every line. > > No. The diagonal contains all the d_nn. No line contains > all the d_nn. The diagonal is longer than every line. The diagonal consists of line indexes, i.e., of the line ends. Therefore it is a subset of the line indexes. > > > > Only if the complete column does exist. But just that is wrong. > > > > If we assume case iia the complete column does exist.. The assumption leads to a contradiction. Therefore the assumption is false. > > > > adding one element to every line does not add a column. > > > So if we start with the same number of lines and columns > > > we do not end with the same number of lines and columns. > > > > That should show you that the assertion of an actually infinite set of > > finite numbers is a self contradiction. > > No. it just shows that the assertion of an actually infinite set of > finite > numbers leads to results that you do not like. However, these > are not contradictory results. These are: 1) The diagonal must be longer than every line. 2) The diagonal cannot be longer than every line. That is in my opinion a contradiction. > > Why does addition of one element yield different results for columns, > > diagonal and lines? > > The columns and the diagonal both contain an infinite initial > segment. No line contains an infinite initial segment. I know that. But you should recognize that this is a contradiction. If not, try to transpose the matrix. > > > > It does (the set of finite segments). Look at the introductory sketch. > > > > The set of finite segments is the same. There is one segment that > is not finite. The set of initial segments does not correspond to > the set of lines. That is true. Therefore there is no infinite set of finite segments, contrary to the assumption. > > > > > The bijection is only valid for finitely many segments (finitely many > > numbers) > The segments count them selves. "Finite" and "finitely many" are the same as the matrix shows. > No. > > The bijection is only valid for finite (not finitely many) segments. > By iia there are an infinitely many finite sements. > > > and lines with finite index (finite numbers). > > By iia there are an infinite number of finite numbers. > > >We cannot give > > an upper bound, but we can exclude an infinite ordinal and cardinal. > > > > Not if you assume the set of all natural numbers, N, exists. The assumption has been assumed. The assumption leads to a contradiction. Therefore the assumption is false. Regards, WM
From: Franziska Neugebauer on 19 Nov 2006 08:26 mueckenh(a)rz.fh-augsburg.de wrote: > Franziska Neugebauer schrieb: [...] >> >> > You need not interpret n as a set (though you can do it). >> >> >> >> In contemporary set theory almost everything is a set. >> > >> > In ZFC everything is a set. Not in every set theory. >> >> I don't want to debate about the axioms of ZFC being sets. The point >> is that you want to talk about "columns" which neither ZFC nor any >> other contemporary set theory is about. > > I think that one can introduce new expressions, examples, and > illustrations if they have been made sufficiently clear to a > correspondent of average IQ. And obviously some increased IQ is necessary to distinguish propositions of set theory from those of "columns". > The reaction of William confirms that my ideas are understandable. So > your reaction does not concern my writings but rather your means of > reception. You may even cite D. Marcus as your witness: ,----[ <MPG.1fc97b8dfd9b203f989941(a)news.rcn.com> ] | I think we'll have more fun if we talk to him on his terms rather than | try to get him to be precise and formal. He doesn't know how to do the | latter. So, he just reverts to repeating himself. `---- >> >> A treatise in which variables ("n") and number symbols ("0", "1", >> >> ...) do not refer to sets is not a treatise _on_ set theory but >> >> a treatise of _application_ of set theory, if ever. >> > >> > Don't mistake set theory with ZF or ZFC. >> >> A theory of _sets_ is not a theory of _columns_. > > Experience has shown that practically all notions used in contemporary > mathematics can be defined, and their mathematical properties derived, > in ZFC. In this sense, the axiomatic set theory serves as a > satisfactory foundation for al[l] other branches of mathematics. > It can describe every mathematical notion --- with the exception of > what a column is? I do not speculate about what is describable and what not. If you want to posit a definition do so! It is curious that you obviously don't like to give precise definitions of certain notions even when you have explicitly been asked for. [...] >> > Sometimes it is necessary to quote. In particular if you are >> > uninformed but nevertheless refuse to take advice from me. >> >> I don't need any advice from you. > > You don't know it. That' s why you would need to learn a lot. Look, > you have learned from me meanwhile, even against your furious > opposition, that omega = |omega| in modern set theory. You do no longer persue your plan to prove a contradiction in modern set theory? > Doesn't this case make you wonder whether there are other things which > you do not yet know but which you could learn from me? > >> My opinion is >> >> A (n e omega & |{0, 1, 2, ..., n}| < |omega|) > > Seems an empty opinion. What is the symbol A refers to? A n (n e omega & |{0, 1, 2, ..., n}| < |omega|) >> >> > As the diagonal is defined to consist of the ends of terms >> >> > of lines, this claim is easy to conradict. >> >> >> >> "ends of terms of lines" is your wording not mine. >> > >> > It is not your wording, because you prefer to veil your >> > inconsistencies, but it is your opinion. >> >> My opinion is >> >> |(d_nn) n e omega| = |omega| > > I know. You say there are less natural numbers than omega, but there > are as many natural numbers as omega. What I say is: The cardinality of the sequence (d_nn) is the cardinality of omega. >> > Can you understand mathematical symbols? Can you understand d_nn >> > <--> n? >> >> Writing down a bunch of symbols does not mean that you created a >> mathematical notation. Rephrase what precisely "There are more >> natural numbers d_nn than natural numbers n." shall mean. > > It means that we have discovered an inconsistency in the assumption > that there were infinitely many finite numbers. We -> You discovered -> invented F. N. -- xyz
From: Franziska Neugebauer on 19 Nov 2006 08:30
mueckenh(a)rz.fh-augsburg.de wrote: > Franziska Neugebauer schrieb: >> >> > The cardinality of omega is omega. >> >> >> >> The cardinality of omega is |omega| not omega. >> > >> > Kunen's "Set Theory" defines |A| to be the least ordinal that can >> > be bijected with A. So, with this definition, |omega| = omega. >> >> You are absolutely right. > > Well learned (after all)! > > Now try to understand the next step: > > If omega exists, then |omega| =/= omega & |omega| = omega. Sorry how did you arrive at |omega| =/= omega (*) ? F. N. -- xyz |